ISS 2017 Statistics Paper-1 Solution: Question 2 (Expected Run Length in Bernoulli Trials)
Continuing our question-by-question walk through the ISS Statistics Paper-1 archive, today we pick up right where the previous post left off and tackle Question 2 of the 2017 paper. This one lives in the probability section and tests whether you can correctly set up and sum an infinite series once you've identified the right distribution.
Quick Summary
Topic:Probability — runs in Bernoulli trials, geometric distribution, expectation of an infinite series
Question Reference:ISS 2017, Statistics Paper-1, Question 2
Correct Answer:Option (b) — 2.5
The Question, As Asked
Consider a sequence of Bernoulli trials with probability of success 1/3 for each trial. Let X denote the length of run of either successes or failures starting with the first trial. Then the value of E(X) is
2
2.5
3
3.5
Step 1: Understanding What a “Run” Means Here
Before touching any formula, make sure the picture in your head is correct. We have an endless sequence of Bernoulli trials — each trial is either a Success (S) or a Failure (F), with P(S) = p = 1/3 and P(F) = q = 1 − p = 2/3.
Now look only at the first trial. Whatever it turns out to be — say it's S — we keep walking forward through the sequence counting how many trials in a row give the same outcome (S, S, S, …) until we hit the first trial that breaks the pattern (the first F). The number of trials counted, including that very first trial, is the random variable X. If the first trial had been F instead, we'd count consecutive F's the same way.
So X is always at least 1 (the first trial itself always counts), and it can in principle be arbitrarily large if we keep getting lucky repeats.
Step 2: The Cleanest Way In — Condition on the First Trial
The trick that makes this problem easy is to split it into two cases based on what the first trial actually is, and handle each case as its own mini-problem.
Case 1: First trial is a Success (probability p).Given this, the run continues for as long as we keep getting S (each with probability p), and it stops the moment we get an F (probability q). So:
P(X = k | first trial is S) = pk−1 · q, for k = 1, 2, 3, …
This is exactly the geometric distribution that counts “number of trials up to and including the first failure,” where each individual trial succeeds (i.e., extends the run) with probability p and fails (i.e., stops the run) with probability q. For a random variable with this exact shape, the mean is a standard result:
E(X | first trial is S) = 1/q
If you haven't memorized that formula, here's where it comes from, shown in full rather than just asserted. We want S = ∑k=1∞ k · pk−1 q. Factor out q: S = q · ∑k=1∞ k pk−1. Now recall the standard power-series identity ∑k=1∞ k xk−1 = 1/(1−x)2 for |x| < 1 (this comes from differentiating the geometric series ∑ xk = 1/(1−x) term by term with respect to x). Plugging in x = p gives ∑ k pk−1 = 1/(1−p)2 = 1/q2. So S = q · 1/q2 = 1/q, confirming the formula.
Case 2: First trial is a Failure (probability q).By exactly the same logic with the roles of S and F swapped, the run continues as long as we get F (probability q each time) and stops at the first S (probability p). So by symmetry:
E(X | first trial is F) = 1/p
Pro Tip: The single most common slip on this question is to average 1/p and 1/q as if they were equally weighted, i.e., writing E(X) = (1/p + 1/q)/2. That's wrong, because the two cases aren't equally likely unless p = q = 0.5. You must weight each conditional expectation by the probability of being in that case — this is just the law of total expectation, E(X) = P(S)·E(X|S) + P(F)·E(X|F), and it's a habit worth drilling until applying it correctly becomes completely automatic rather than something you have to stop and think about mid-exam.
Step 3: Combine the Two Cases Using the Law of Total Expectation
Now put the pieces together:
E(X) = P(first trial is S) · E(X | first trial is S) + P(first trial is F) · E(X | first trial is F)
E(X) = p · (1/q) + q · (1/p) = p/q + q/p
Substitute p = 1/3 and q = 2/3:
p/q = (1/3) ÷ (2/3) = 1/2
q/p = (2/3) ÷ (1/3) = 2
E(X) = 1/2 + 2 = 2.5
Step 4: A Second, Independent Route (Cross-Check)
It's always reassuring to verify an answer using a completely different line of attack rather than just re-reading the same algebra. Here, instead of conditioning, we can write down the unconditional probability mass function of X directly and sum it from scratch.
For X = k to happen, the first k trials must all be the same outcome (either all S, with probability pk, or all F, with probability qk), and then trial k+1 must differ from that run (probability q if the run was S's, probability p if the run was F's). So:
P(X = k) = pkq + qkp, for k = 1, 2, 3, …
As a sanity check, this should sum to 1 over all k. Using ∑k=1∞ pk = p/(1−p) = p/q, the first term sums to q · p/q = p; by the symmetric argument the second term sums to q. Together p + q = 1, so the distribution is valid.
Now compute E(X) = ∑k=1∞ k(pkq + qkp) = q·∑k pk + p·∑k qk. Using the same differentiated-geometric-series identity as before (this time ∑k=1∞ k xk = x/(1−x)2), we get ∑k pk = p/q2 and ∑k qk = q/p2. So:
E(X) = q · (p/q2) + p · (q/p2) = p/q + q/p = 1/2 + 2 = 2.5
Both methods agree exactly, which is a strong indication there's no arithmetic slip hiding anywhere.
Final Answer and Verification Against the Official Key
Our fully worked-out answer is E(X) =2.5, which corresponds to option (b). Checking against the official ISS 2017 Statistics Paper-1 answer key (Series A, since that matches this test booklet) confirms the same: option (b) is listed as correct. So our derivation and the official key are in full agreement — no discrepancy to flag here.
Why This Question Matters
This question is a good small laboratory for two skills that show up constantly in the ISS Statistics papers: recognizing when a geometric distribution is hiding inside a word problem, and correctly applying the law of total expectation when a random variable's behaviour splits naturally into cases. The run-length idea itself also reappears in run tests for randomness, reliability analysis, and time-series pattern detection, so understanding it from first principles (rather than memorizing “the answer is p/q + q/p”) pays off well beyond this one MCQ. The way this derivation is usually taught is to insist on writing out the conditioning step explicitly every single time, even when it feels obvious, because that habit is exactly what prevents the unequal-weighting mistake described in the Pro Tip above under exam pressure.
Frequently Asked Questions
What exactly is a “run” in a sequence of Bernoulli trials?
A run is an unbroken stretch of identical outcomes. In this problem, the run we care about always starts at the very first trial and keeps extending for as long as subsequent trials match that first outcome, stopping the instant a different outcome appears.
Why is the conditional distribution of X geometric?
Once you fix what the first trial's outcome is, say Success, every later trial independently either repeats that outcome (extending the run) or breaks it (ending the run). Counting the number of trials until the pattern breaks is precisely the definition of a geometric random variable, just phrased in terms of “matching” and “breaking” instead of the usual “failure” and “success” labels.
Is there a general formula I can remember for this type of problem?
Yes: for a Bernoulli sequence with success probability p and failure probability q = 1 − p, the expected length of the run starting at the first trial is always E(X) = p/q + q/p. You can plug in any value of p directly into this formula once you've derived it once and understood where it comes from.
What would the answer be if the trials were fair, i.e., p = q = 0.5?
Plugging p = q = 0.5 into E(X) = p/q + q/p gives 1 + 1 = 2. This makes intuitive sense: with a fair coin there's no asymmetry between success-runs and failure-runs, so the expected run length settles at a smaller, “balanced” value compared to the 2.5 we got with the biased 1/3-2/3 split in this question.
What is the most common mistake students make on this type of question?
The most frequent error is averaging the two conditional expectations 1/p and 1/q without weighting them by how likely each starting case is — that is, forgetting to multiply by p and q respectively before adding. A second common slip is misapplying the power series formula for ∑kxk-1 versus ∑kxk, which are off by one factor of x from each other.
Does this kind of run-based probability question appear often in ISS Paper-1?
Questions built around runs, waiting times, and geometric/negative-binomial reasoning show up fairly regularly across ISS Statistics Paper-1 papers, since they test both distributional intuition and series-summation skill in one go. Drilling a handful of these until the conditioning step becomes second nature is one of the more efficient ways to lock in marks in this area.
If anything in this derivation felt unclear, or if you solved it differently and want a second opinion, drop your question in the comments — and if you know a fellow ISS aspirant working through the same papers, feel free to pass this along to them too.
Comments